`Ph-underset(underset(Cl)(|))(CH)-CH_(2)-Cl underset(underset(("iii") H^(+))(("ii")NaNH_(2)))overset(("i")Alc.kOH)(rarr) A overset("Cold alk." KMnO_(4))(rarr)B underset("warm")overset("NaOH")(rarr)C`
The product 'C' eill be

Options

A. `Ph-overset(OH)overset(|)CH-COO`
B. `Ph-underset(underset(OH)(|))CH-underset(underset(OH)(|))CH_(2)`
C. `Ph-CH_(2)-COO^(-)`
D. `Ph-overset(overset(O)(||))C-underset(underset(OH)(|))CH_(2)`

Explanation

To solve the given question step by step, we will analyze the reactions and transformations that occur to the starting compound through the specified reagents.

### Step 1: Identify the Starting Compound
The starting compound is a phenyl group connected to a carbon chain with a chlorine atom (Cl) attached to it. The structure can be represented as:
\[ \text{Ph-CH(CH}_2\text{-Cl)} \]

### Step 2: Treatment with Alcoholic KOH or NaNH2
When the compound is treated with alcoholic KOH or sodamide (NaNH2), both act as dehydrohalogenating agents. They will remove a hydrogen atom and a chlorine atom, resulting in the formation of a double bond.

The reaction can be summarized as:
\[ \text{Ph-CH(CH}_2\text{-Cl)} \xrightarrow{\text{alc. KOH or NaNH}_2} \text{Ph-CH=CH-Cl} + \text{HCl} \]

### Step 3: Forming Product A
After the elimination of HCl, we obtain product A:
\[ \text{A: Ph-CH=CH-Cl} \]

### Step 4: Treatment with Cold Alkaline KMnO4
Next, product A is treated with cold alkaline KMnO4, which is known as Baeyer's reagent. This reagent adds hydroxyl (OH) groups across the double bond.

The reaction can be represented as:
\[ \text{Ph-CH=CH-Cl} \xrightarrow{\text{cold KMnO}_4} \text{Ph-CH(OH)-CH(OH)-Cl} \]

### Step 5: Forming Product B
After the reaction with KMnO4, we get product B:
\[ \text{B: Ph-CH(OH)-CH(OH)-Cl} \]

### Step 6: Treatment with Warm NaOH
Finally, product B is treated with warm sodium hydroxide (NaOH). The strong base will facilitate the substitution reaction, leading to the formation of a carboxylic acid.

The reaction can be summarized as:
\[ \text{Ph-CH(OH)-CH(OH)-Cl} \xrightarrow{\text{warm NaOH}} \text{Ph-COOH} + \text{NaCl} + \text{H}_2\text{O} \]

### Step 7: Forming Product C
After the reaction with warm NaOH, we obtain product C:
\[ \text{C: Ph-COOH} \]

### Final Answer
The product 'C' will be:
\[ \text{C: Ph-COOH (benzoic acid)} \]

---